anonymous
  • anonymous
please help! find the absolute max and min of f(x)=x^3-12x on [0,4]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
answer : 2
anonymous
  • anonymous
y(dx/dy) = 3x^2 -12 getting the critical numbers: 3x^2 - 12 = 0 3x^2 = 12 x^2 = 12/3 x^2 = 4 x = +-2 substitute it to the original value y1=(2)^3-12(2) y1=-16 y2=(-2)^3-12(-2) y2=16 get the y value of the interval y3 = (0)^3 - 12(0) y3 =0 y4 = (4)^3 - 12(4) y4 = 16 biggest is 16 now we know that abs max is 16 at x = 4 and x = -2 smalles is -16 so abs min is -16 at x=2
anonymous
  • anonymous
find the derivative of f(x) and solve it when it's equal to 0 f'(x)=3x^2-12 \[3x^2-12=0 \implies 3x^2=12 \implies x^2=4\] x=2 or x=-2 but -2 is not in [0,4] all you have to do now is to check the values of the function at x=2, and at the border points x=0, x=4 the maximum among them is the maximum in the interval, and same is applied to the minimum

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
oh yea i forgot x= -2 cant be considered~~ but nevertheless 16 is still the abs maximum ^^
anonymous
  • anonymous
f(2)=2^3-12(2)=8-24=-16 f(0)=0 f(4)=4^3-12(4)=16 so the absolute maximum is 16 at x=4 and the absolute minimum is -16 at x=2

Looking for something else?

Not the answer you are looking for? Search for more explanations.